Non-Homogeneous Linear Equations

Applying the Rouché-Capelli theorem (§1.5) to non-homogeneous systems, four representative cases illustrate every possible outcome:

Unique solution. For a $4\times3$ system with $\rho(A)=\rho([A|B])=3=$ number of unknowns, back substitution from the echelon form gives one solution -- consistent, unique.

One-parameter family. If $\rho(A)=\rho([A|B])=2<3$, the echelon form leaves one equation "used up" (a genuine $0=0$ row), so one unknown is fixed arbitrarily as a parameter $t$ and the other two follow by back substitution -- consistent, infinitely many solutions forming a one-parameter family.

Two-parameter family. If $\rho(A)=\rho([A|B])=1<3$, only one genuinely independent equation survives; two unknowns are fixed arbitrarily as parameters $s,t$ and the third follows -- a two-parameter family.

Inconsistent. If $\rho(A)<\rho([A|B])$ (e.g. $\rho(A)=3,\ \rho([A|B])=4$), the echelon form of $[A|B]$ contains a row reading $0=c$ for some $c\ne0$ -- a flat contradiction, so no solution exists.

Standing rule (with $n$ = number of unknowns):

1. $\rho(A)=\rho([A|B])=n$ $\Rightarrow$ consistent, unique solution.
2. $\rho(A)=\rho([A|B])=n-k<n$ $\Rightarrow$ consistent, infinitely many solutions forming a $k$-parameter family. (For 3 unknowns: $\rho=2$ gives a one-parameter family; $\rho=1$ gives a two-parameter family.)
3. $\rho(A)\ne\rho([A|B])$ $\Rightarrow$ inconsistent, no solution. …